Bruh ima be honest with you idek
<u>Answer:
</u>
Cat has 2.02 seconds to right itself.
<u>Explanation:
</u>
Initial height of cat from ground = 20 meter.
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of cat in vertical direction = 0 m/s, acceleration = acceleration due to gravity = 9.8
, we need to calculate time when s = 20 meter.
Substituting
So, cat has 2.02 seconds to right itself.
Answer:
15.065ft
Explanation:
To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.
By definition the drag force is expressed as:
![F_D = -\frac{1}{2}\rho V^2 C_d A](https://tex.z-dn.net/?f=F_D%20%3D%20-%5Cfrac%7B1%7D%7B2%7D%5Crho%20V%5E2%20C_d%20A)
Where
is the density of the flow
V = Velocity
= Drag coefficient
A = Area
For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3
For second Newton's Law the Force is also defined as,
![F=ma=m\frac{dV}{dt}](https://tex.z-dn.net/?f=F%3Dma%3Dm%5Cfrac%7BdV%7D%7Bdt%7D)
Equating both equations we have:
![m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A](https://tex.z-dn.net/?f=m%5Cfrac%7BdV%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Crho%20V%5E2%20C_d%20A)
![m(dV)=-\frac{1}{2}\rho C_d A (dt)](https://tex.z-dn.net/?f=m%28dV%29%3D-%5Cfrac%7B1%7D%7B2%7D%5Crho%20C_d%20A%20%28dt%29)
![\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BV%5E2%20%7D%28dV%29%3D-%5Cfrac%7B1%7D%7B2m%7D%5Crho%20C_d%20A%20%28dt%29)
Integrating
![\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7B1%7D%7BV%5E2%20%7D%28dV%29%3D%20-%20%5Cint%5Cfrac%7B1%7D%7B2m%7D%5Crho%20C_d%20A%20%28dt%29)
![-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7BV%7D%5Cbig%7C%5E%7BV_f%7D_%7BV_i%7D%3D%5Cfrac%7B1%7D%7B2m%7D%28%5Crho%29C_d%20%28%5Cpi%20r%5E2%29%20%5CDelta%20t)
Here,
![V_f = 60mph = 26.82m/s](https://tex.z-dn.net/?f=V_f%20%3D%2060mph%20%3D%2026.82m%2Fs)
![V_i = 120.7m/s](https://tex.z-dn.net/?f=V_i%20%3D%20120.7m%2Fs)
![m= 1600lbf = 725.747Kg](https://tex.z-dn.net/?f=m%3D%201600lbf%20%3D%20725.747Kg)
![\rho = 1.21 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%201.21%20kg%2Fm%5E3)
![C_d = 0.3](https://tex.z-dn.net/?f=C_d%20%3D%200.3)
![\Delta t=7s](https://tex.z-dn.net/?f=%5CDelta%20t%3D7s)
Replacing:
![\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B26.82%7D%2B%5Cfrac%7B1%7D%7B120.7%7D%20%3D%20%5Cfrac%7B1%7D%7B2%28725.747%29%7D%281.21%29%280.3%29%28%5Cpi%20r%5E2%29%20%287%29)
![-0.029 = -5.4997r^2](https://tex.z-dn.net/?f=-0.029%20%3D%20-5.4997r%5E2)
![r = 2.2963m](https://tex.z-dn.net/?f=r%20%3D%202.2963m)
![d= r*2 = 4.592m \approx 15.065ft](https://tex.z-dn.net/?f=d%3D%20r%2A2%20%3D%204.592m%20%5Capprox%2015.065ft)
Answer:
Explanation:
Maximum force of friction possible = μmg
= .65 x 3.8 x 9.8
= 24.2 N
u = 72 x 1000 / 60 x 60
= 20 m /s
v² = u² - 2as
a = 20 x 20 / (2 x 30)
= 6.67 m / s²
force acting on it
= 3.8 x 6.67
= 25.346 N
Friction force possible is less .
So friction will not be able to prevent its slippage
It will slip off .
Force = (mass) x (acceleration)
If the full 15N is pointing parallel to the ground,
then
15 N = (58 kg) x (acceleration).
Divide each side
by 58 kg: Acceleration = 15 N / 58 kg
= (15 kg-m/s²) / (58 kg)
= (15/58) (kg-m/kg-s²)
= 0.26 m/s² .